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# Find values of a and b such that the given limit exists

Find values for the real constants and so that the following limit equation holds:

Incomplete.

# Find values of constants a and b such that the given improper integral converges

Find values for real numbers and so that the following improper integral equation holds:

First, we have

In order for this to converge we must have the coefficient of in the numerator equal to 0; hence, we must have . Now, to solve the integral equation we have

But, by assumption this integral equals 1, so we have

Since we then have

# Find a value of C so that the given improper integral converges

Find a value for the constant so that the improper integral

converges and find the value of the integral in this case.

First, we have

For this to converge we must have the coefficient of the in the numerator going to 0 as (otherwise the integral will converge by limit comparison with ). Hence,

Now, we need to evaluate the integral. Incomplete.

# Find a value of the constant C so that the given improper integral converges

Find the value of so that the improper integral

converges, and compute the value of the integral in this case.

First, we compute the value of the constant ,

This integral will diverge by comparison with if the coefficient of the term in the numerator is not equal to zero. Hence, we must have

Next, we evaluate the integral with ,

# Find a constant so the given improper integral converges

Find a value for the constant so that the improper integral

converges. Find the value of the integral for this value of .

First, we compute the value of ,

Since this integral will diverge by limit comparison with for any nonzero coefficient of the term in the numerator we must have

Next, we evaluate the integral

# Test the improper integral ∫ 1 / x (log x)s for convergence

Test the following improper integral for convergence:

The integral converges if and diverges for .

Proof. We can compute this directly. We evaluate the indefinite integral (using the substitution ) for :

But then the limit

is finite for and diverges for (since as and so the limit diverges when and converges for ). Hence, the integral

converges for and diverges for . In the case that the indefinite integral is

and as , so the integral divers for as well. Therefore, we have the integral converges for and diverges for

# Test the improper integral ∫ 1 / (x1/2 log x) for convergence

Test the following improper integral for convergence:

The integral diverges.

Proof. First, we show that for all . To do this let

This derivative is 0 at and is less than 0 for and greater than 0 for . Hence, has a minimum at . But, (since implies ). So, has a minimum at and is positive there; thus, it is positive for all , or

So, since we know

But then, consider the limit

Therefore, by the limit comparison test (Theorem 10.25), the convergence of would imply the convergence of (for any ), but we know by Example 5 that this integral diverges. Hence, we must also have the divergence of

# Test the improper integral ∫ log x / (1-x) for convergence

Test the following improper integral for convergence:

The integral converges.

Proof. First, we write

For the first integral we know for all we have ; hence,

Then, the integral

(We know by L’Hopital’s, writing or by Example 2 on page 302 of Apostol.) Hence,

converges by the comparison theorem (Theorem 10.24 on page 418 of Apostol).

For the second integral, we use the expansion of about ,

Then we have

But this integral converges since it has no singularities.

Thus, we have established the convergence of

# Test the improper integral ∫ log x / x1/2 for convergence

Test the following improper integral for convergence:

The integral converges.

Proof. We can compute the value of the improper integral directly. First, we can evaluate the indefinite integral using integration by parts with

Therefore,

So, to evaluate the improper integral we take the limit,

# Test the improper integral ∫ e-x1/2 / x1/2 for convergence

Test the following improper integral for convergence:

The integral converges.

Proof. We can compute this integral directly. First, we evaluate the indefinite integral using the substitution , .

Now, we have discontinuities at both limits of integration so we evaluate by taking two limits,